MHB What Are the Solutions for the Exponential Equation \( x^{(y^z)} = (x^y)^z \)?

[anemone]

Here is this week's POTW:

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Determine all triples $(x,\,y,\,z)$ of positive integers with $x^{(y^z)}=\left(x^y\right)^z$.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[anemone]

Congratulations to kaliprasad for his correct solution!:)

Here's the suggested model solution:

Spoiler

The equation $x^{(y^z)}=\left(x^y\right)^z$ is equivalent to the equation $x^{(y^z)}=x^yz$. We examine a number of different cases.

Case 1: $x=1$
Then the equation is true regardless of the values of $y$ and $z$. Therefore, $(1,\,y,\,z)$ is a solution for all positive integers $y$ and $z$.

Case 2: $x>1$
In this case, $x^{(y^z)}=\left(x^y\right)^z$ is equivalent to $y^z=yz$, which is equivalent to $y^{z-1}=z$ since $y>0$. We consider subcases where $z=1,\,z=2$ and $z>2$.

Subcase 2(a): $x>1$ and $z=1$.
If $z=1$, then we have $y^0=1$, which is true for all positive integers $y$. Therefore $(x,\,y,\,1)$ is a solution for all positive integers $x>1$ and all positive integers $y$.

Subcase 2(b): $x>1$ and $z=2$.
If $z=2$, then the equation $y^{z-1}=z$ becomes $y=2$. Therefore $(x,\,2,\,2)$ is a solution for all positive integers $x>1$.

Subcase 2(c): $x>1$ and $z>2$.
If $z>2$, then $y$ cannot equal 1, so $y\ge 2$. Using the fact that $2^{z-1}>z$ for $z\ge 3$, we see that that $y^{z-1}\ge 2^{z-1}>z$, so $y^{z-1}=z$ has so solutions in this case.

In conclusion, the solutions are all triples $(x\,y,\,z)$ of positive integers with
(i) $x=1$ or
(ii) $x>1$, $z=1$ or
(iii) $x>1$ and $y=z=2$.

To finish, we must show that $2^{z-1}>z$ for all positive integers $z\ge 3$. We prove this by mathematical induction on $z$.

If $z=3$, the inequality becomes $4=2^2>3$, which is true.

Suppose that $2^{z-1}>z$ for $z=k$ for some positive integer $k\ge 3$.

Consider $z=k+1$. Since $2^{k-1}>k$ by the induction hypothesis, then $2^k=2\cdot 2^{k-1}>2k$. Since $k\ge 3$, then $2k>k+1$ so $2^k>k+1$ or $2^{k+1)-1}>k+1$ as required. This completes the proof by induction.